Supplementary Material for
“Interacting Reinforced Stochastic Processes:
Statistical Inference based on the Weighted Empirical Means”
GIACOMO ALETTI1,∗ IRENE CRIMALDI2,† and ANDREA GHIGLIETTI3
1ADAMSS Center, Universit`a degli Studi di Milano, Milan, Italy E-mail:giacomo.aletti@unimi.it
2IMT School for Advanced Studies, Lucca, Italy E-mail:irene.crimaldi@imtlucca.it
3LivaNova, Milan, Italy
E-mail:Andrea.Ghiglietti@guest.unimi.it
In this document we collect some technical results and computations necessary for the proof of (Aletti, Crimaldi and Ghiglietti,2019, Lemma5.1) and we give the entries of the matrices Sγ11, Sγ12and Sγ22introduced in (Aletti, Crimaldi and Ghiglietti, 2019, Theorem5.1). Therefore, the notation and the assumptions used here are the same as those used in that paper.
A. Technical details for the proof of (Aletti, Crimaldi and Ghiglietti, 2019, Lemma 5.1)
In all the sequel, given (zn)n, (zn0)n two sequences of complex numbers, the notation zn = O(zn0) means
|zn| ≤ C|z0n| for a suitable constant C > 0 and n large enough. Moreover, if zn0 6= 0, the notation zn∼ zzn0 with z ∈ C \ {0} means limnzn/zn0 = z and, finally, the notation zn= o(z0n) means limnzn/zn0 = 0.
Given 1/2 < δ ≤ 1, x = ax+ i bx∈ C with ax > 0 and an integer m0 ≥ 2 such that axm−δ < 1 for all m ≥ m0, let us set
pδn(x) :=
n
Y
m=m0
1 − x
mδ
for n ≥ m0. (A.1)
A.1. Some technical results
We first recall the following result, which has been proved inAletti, Crimaldi and Ghiglietti(2017).
Lemma A.1. (Aletti, Crimaldi and Ghiglietti,2017, Lemma A.4) We have
|pδn(x)| = (O
exp
−axn1−δ1−δ
for 1/2 < δ < 1 O (n−ax) for δ = 1
(A.2)
∗Member of “Gruppo Nazionale per il Calcolo Scientifico (GNCS)” of the Italian Institute “Istituto Nazionale di Alta Matematica (INdAM)”
†Member of “Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA)” of the Italian Institute “Istituto Nazionale di Alta Matematica (INdAM)”
1
and
|pδn(x)−1| = (O
exp axn1−δ
1−δ
for 1/2 < δ < 1 O (nax) for δ = 1 .
(A.3)
Therefore , if we set
Fk+1,nδ (x) := pδn(x)
pδk(x) for m0≤ k ≤ n , (A.4)
we have
|Fk+1,nδ (x)| =
O
exp
ax
1−δ(k1−δ− n1−δ)
for 1/2 < δ < 1 O
k n
ax
for δ = 1.
(A.5)
Now, we prove two other results.
Lemma A.2. Given β > 1 and e > 0, we have
n
X
k=m0
1
kβ|Fk+1,nδ (x)|e =
O n−(β−δ)
if 1/2 < δ < 1,
O (n−eax) if δ = 1 and eax< β − 1, O n−(β−1)ln(n)
if δ = 1 and eax= β − 1, O n−(β−1)
if δ = 1 and eax> β − 1.
(A.6)
Proof. The desired relations immediately follows from (A.5) using the well-known relation
n
X
k=1
1 k1−a =
O(1) for a < 0,
ln(n) + d + O(n−1) = ln(n) + O(1) for a = 0, a−1na+ O(1) for 0 < a ≤ 1,
a−1na+ O(na−1) for a > 1,
(A.7)
where d is the Euler-Mascheroni constant, and the relation
n
X
k=1
exp(akb/b)
kβ = O
Z n 1
exp(atb/b) tβ dt
= O exp(atb/b) atb+β−1
n
1
+(b + β − 1) a
Z n 1
exp(atb/b) tβ+b dt
= O exp(anb/b) nb+β−1
for a > 0, b > 0, β > 1.
(A.8)
Indeed, for the case δ = 1, it is enough to apply (A.7) with a = eax− (β − 1); while, for the case 1/2 < δ < 1, it is enough to apply (A.8) with a = eax, b = 1 − δ and β.
The following lemma extends (Aletti, Crimaldi and Ghiglietti,2017, Lemma A.5).
Lemma A.3. Given 1/2 < δ1≤ δ2≤ 1, β > δ1 and x1, x2∈ C with Re(x1) > 0, Re(x2) > 0, let m0≥ 2 be an integer such that max{Re(x1), Re(x2)}m−δ1 < 1 for all m ≥ m0. Then we have
limn nβ−δ1
n
X
k=m0
k−βFk+1,nδ1 (x1)Fk+1,nδ2 (x2) =
1
x1+x2 if 1/2 < δ1= δ2< 1,
1
x1+x2−β+1 if δ1= δ2= 1 and Re(x1+ x2) > β − 1,
1
x1 if 1/2 < δ1< δ2≤ 1.
(A.9)
Proof. Let us start with observing that, in each considered case, relation (A.2) implies
limn nβ−δ1|pδn1(x1)| |pδn2(x2)| = 0. (A.10) Indeed, in particular, when δ1= δ2= 1 we have the additional condition Re(x1+ x2) > β − 1.
Now, fix k ≥ 2 and let us set η := β − δ1and `δn(x) := 1/pδn(x) and define the following quantity Dk := 1
kη`δk1(x1)`δk2(x2) − 1
(k − 1)η`δk−11 (x1)`δk−12 (x2)
= 1
kη − 1 (k − 1)η
`δk−11 (x1)`δk−12 (x2) + 1 kη
`δk1(x1)`δk2(x2) − `δk−11 (x1)`δk−12 (x2)
= `δk1(x1)`δk2(x2)
"
1
kη − 1 (k − 1)η
`δk−11 (x1)`δk−12 (x2)
`δk1(x1)`δk2(x2) + 1
kη 1 − `δk−11 (x1)`δk−12 (x2)
`δk1(x1)`δk2(x2)
!#
. Then, we observe the following:
1
kη − 1 (k − 1)η
= − η k1+η + O
1 k2+η
for k → +∞ (A.11)
and
`δk−11 (x1)`δk−12 (x2)
`δk1(x1)`δk2(x2) = 1 − x1
kδ1
1 − x2
kδ2
= 1 + x1x2
k(δ1+δ2)− x1
kδ1 − x2
kδ2. (A.12) Now, by using (A.11) and (A.12) in the above expression of Dk, we have for k → +∞
Dk = `δk1(x1)`δk2(x2)
− η
kη+1 + O(1/kη+2)
1 + x1x2
k(δ1+δ2) − x1
kδ1 − x2
kδ2
+ 1
kη
− x1x2
k(δ1+δ2) + x1
kδ1 + x2
kδ2
=
(`δk(x1)`kδ(x2)x1+x2
kη+δ −kη+1η + o(1/kη+δ)
if δ1= δ2= δ,
`δk1(x1)`δk2(x2) x1
kη+δ1 + o(1/kη+δ1)
if δ1< δ2
=
`δk(x1)`δk(x2)xk1η+δ+x2 + o(`δk(x1)`δk(x2)/kη+δ) if δ1= δ2= δ < 1,
`1k(x1)`1k(x2)x1+xkη+12−η + o(`k1(x1)`1k(x2)/kη+1) if δ1= δ2= 1 and Re(x1+ x2) > η,
`kδ1(x1)`δk2(x2)kη+δ1x1 + o(`kδ1(x1)`δk2(x2)/kη+δ1) if δ1< δ2. that is
Dk ∼
x1+x2
kη+δ `δk(x1)`kδ(x2) if 1/2 < δ1= δ2= δ < 1,
x1+x2−η
kη+1 `1k(x1)`1k(x2) if δ1= δ2= 1 and Re(x1+ x2) > η,
x1
kη+δ1 `δk1(x1)`kδ2(x2) if 1/2 < δ1< δ2≤ 1.
(A.13)
Now, following the same arguments used in the proof of (Aletti, Crimaldi and Ghiglietti,2017, Lemma A.5), in order to conclude, we apply (Aletti, Crimaldi and Ghiglietti,2017, Corollary A.2) with
zn = Dn, vn= nηpδn1(x1)pδn2(x2), wn= `n,1`n,2 nη+δ1Dn
, w =
1
x1+x2 if 1/2 < δ1= δ2= δ < 1
1
x1+x2−η if δ1= δ2= 1 and Re(x1+ x2) > η
1
x1 if 1/2 < δ1< δ2≤ 1.
Indeed, limnvn= 0 by (A.10), limnwn = w 6= 0 by (A.13), limn vn
n
X
k=m0
zk= lim
n nηpδn1(x1)pδn2(x2)
n
X
k=m0
Dk
= lim
n nηpδn1(x1)pδ22(x2) `δn1(x1)`δn2(x2) nη −`δm1
0−1(x1)`δm2
0−1(x2) (m0− 1)η
!
= 1 by (A.10) and zn0 = znwn= r2n`n,1`n,2.
A.2. Analytic expression of A
jk+1,n−1with j ≥ 2
Let us recall the definition of the following quantities for j ≥ 2:
Ajk+1,n−1=
n−1
Y
m=k+1
(I − DQ,j,m) , where DQ,j,n=
brn−1(1 − λj) 0
−λjhn(λj) qbn,n
with the function hn defined in (Aletti, Crimaldi and Ghiglietti, 2019, Equation (5.10)). The aim of this section is to compute the product above and so finding the following useful expression of Ajk+1,n−1:
Ajk+1,n−1=
Fk+1,n−1γ (cαj) 0 λjGk+1,n−1(cαj, q) Fk+1,n−1ν (q)
, where
Gk+1,n−1(cαj, q) =
n−1
X
l=k+1
Fl+1,n−1γ (cαj)hl(λj)Fk+1,l−1ν (q).
It is straightforward to see that [Ajk+1,n−1]21= 0,
[Ajk+1,n−1]11 =
n−1
Y
m=k+1
(1 −rbm−1(1 − λj)) = Fk+1,n−1γ (cαj),
[Ajk+1,n−1]22 =
n−1
Y
m=k+1
(1 −bqm,m) = Fk+1,n−1ν (q),
while it is not immediate to determine [Ajk+1,n−1]12. To this end, let us set xn−1:= [Ajk+1,n−1]21and observe that, since Ajk+1,n−1= Ajk+1,n−2(I − DQ,j,n−1) and xk+1= λjhk+1(λj), we have that
xn−1= xn−2(1 −brn−1(1 − λj)) + [Ajk+1,n−2]22λjhn−1(λj)
= xn−2Fn−1,n−1γ (cαj) + Fk+1,n−2ν (q)λjhn−1(λj)
= xn−3Fn−2,n−1γ (cαj) + Fk+1,n−3ν (q)λjhn−2(λj)Fn−1,n−1γ (cαj) + Fk+1,n−2ν (q)λjhn−1(λj)
= . . .
= xk+1Fk+2,n−1γ (cαj) +
n−2
X
l=k+2
Fk+1,l−1ν (q)λjhl(λj)Fl+1,n−1γ (cαj) + Fk+1,n−2ν (q)λjhn−1(λj)
=
n−1
X
l=k+1
Fk+1,l−1ν (q)λjhl(λj)Fl+1,n−1γ (cαj)
= λjGk+1,n−1(cαj, q) .
A.3. Asymptotic behavior of G
k+1,n−1(x, q)
Let us recall the definition
Gk+1,n−1(x, q) :=
n−1
X
l=k+1
Fl+1,n−1γ (x)hl(1 − c−1x)Fk+1,l−1ν (q).
Here we prove the following result:
Lemma A.4. When ν = γ, we have for x ∈ C \ {0}
Gk+1,n−1(x, q) =
q x−q
Fk+1,n−1γ (q) − Fk+1,n−1γ (x)
if x 6= q,
q
1−γFk+1,n−1γ (q)(n − 1)1−γ− (k + 1)1−γ + O(Fk+1,n−1γ (q)) if x = q and 1/2 < γ < 1, qFk+1,n−1γ (q) ln
n−1 k+1
+ O(k−1Fk+1,n−1γ (q)) if x = q and γ = 1.
(A.14) When ν 6= γ, we have for x ∈ C \ {0}
Gk+1,n−1(x, q) = C(x, q) Fk+1,n−1ν (q)
(n − 1)µ −Fk+1,n−1γ (x) kµ
!
+ O |Fk+1,n−1ν (q)|
n2µ +|Fk+1,n−1γ (x)|
k2µ
!
, (A.15)
where µ := |γ − ν| and
C(x, q) :=
(−xq if ν < γ,
q
x if γ < ν.
Proof. Recalling the definition (A.4), we can write
Gk+1,n−1(x, q) =
n−1
X
l=k+1
Fl+1,n−1γ (x)hl(1 − c−1x)Fk+1,l−1ν (q) =
n−1
X
l=k+1
pγn−1(x)
pγl(x) hl(1 − c−1x)pνl−1(q) pνk(q)
= pγn−1(x) pνk(q)
n−1
X
l=k+1
hl(1 − c−1x)
(1 −qbl,l) Xl, where Xl := pνl(q) pγl(x).
(A.16)
Moreover, recalling the definition of the function hl (see (Aletti, Crimaldi and Ghiglietti, 2019, Equation (5.10))), we have for x 6= 0
hl(1 − c−1x) = (
brl−1c−1x = xl−γ if ν < γ, bql,l= ql−ν if ν ≥ γ.
Let us start with the case ν = γ. In this case, we have
∆Xl:= Xl− Xl−1=
1 −Xl−1
Xl
Xl= x − q q
q lγ(1 − ql−γ)
Xl
= x − q q
qbl,l 1 −qbl,l
Xl=x − q q
hl(1 − c−1x) 1 −qbl,l
Xl. It follows that
x − q q
n−1
X
l=k+1
hl(1 − c−1x) 1 −qbl,l
Xl= Xn−1− Xk. Since
pγn−1(x)
pνk(q) Xn−1 = pγn−1(x) pνk(q)
pνn−1(q)
pγn−1(x) = Fk+1,n−1ν (q) and pγn−1(x)
pνk(q) Xk = pγn−1(x) pνk(q)
pνk(q)
pγk(x) = Fk+1,n−1γ (x),
(A.17)
we find by (A.16)
x − q
q Gk+1,n−1(x, q) =
Fk+1,n−1γ (q) − Fk+1,n−1γ (x)
and so for x 6= q we get
Gk+1,n−1(x, q) = q x − q
Fk+1,n−1γ (q) − Fk+1,n−1γ (x) .
When ν = γ and x = q, we have Xl= 1 and so we obtain (by (A.16) together with (A.7))
Gk+1,n−1(x, q) = qFk+1,n−1γ (q)
n−1
X
l=k+1
1
lγ(1 − ql−γ) = qFk+1,n−1γ (q)
n−1
X
l=k+1
1 lγ + O
X
l≥k+1
l−2γ
= qFk+1,n−1γ (q)
((n−1)1−γ
1−γ −(k+1)1−γ1−γ + O(1) + O(k−(2γ−1)) if 1/2 < γ < 1 ln(n − 1) − ln(k + 1) + O(n−1) + O(k−1) if γ = 1, which implies the two different asymptotic behavior in (A.14) according to the value of γ.
Now, let us consider the case ν 6= γ and introduce the sequence {yl; l ≥ 1} defined as yl := l−µ, with µ = |γ − ν|. Then, we have
ylXl− yl−1Xl−1= ∆ylXl+ yl−1∆Xl= 1
lµ − 1 (l − 1)µ
Xl+ 1 lµ + O
1 l1+µ
∆Xl
= −µ l1+µ + O
1 l2+µ
Xl+ 1 lµ + O
1 l1+µ
∆Xl, where
∆Xl:= Xl− Xl−1=
1 −Xl−1
Xl
Xl= RlXl
with
Rl:=
1 −Xl−1
Xl
= xl−γ− ql−ν
1 − ql−ν = brl−1c−1x −bql,l
1 −bql,l = O
1
lmin{γ,ν}
. Taking into account that µ + min{γ, ν} < 1 + µ for ν 6= γ, we obtain that
ylXl− yl−1Xl−1= Rl lµ + O
1 l1+µ
Xl= K(x, q)hl(1 − c−1x)
(1 −qbl,l) Xl+ QlXl, where
K(x, q) :=
−q
x 1{ν<γ}+ x q
1{ν>γ}= C(x, q)−1 and
Ql:=
(xl−(2γ−ν)
1−bql,l + O(l−(1+µ)) if ν < γ,
−ql−(2ν−γ)1−
bql,l + O(l−(1+µ)) if ν > γ.
Note that Ql∼ κl−(2µ+min{γ,ν}) with a suitable κ 6= 0. The above expression implies that Xn−1
(n − 1)µ −Xk
kµ =
n−1
X
l=k+1
(ylXl− yl−1Xl−1) = K(x, q)
n−1
X
l=k+1
hl(1 − c−1x) (1 −qbl,l) Xl+
n−1
X
l=k+1
QlXl. (A.18)
With similar computations, setting
R∗l := 1 − |Xl−1|
|Xl| = |1 − ql−ν| − |1 − xl−γ|
|1 − ql−ν|
and taking into account that R∗ll−2µ∼ κ0l−(2µ+min{γ,ν})with a suitable κ0 6= 0 and min{γ, ν} < 1 for ν 6= γ, we find
|Xl|
l2µ − |Xl−1|
(l − 1)2µ = R∗l l2µ + O
1 l1+2µ
|Xl| = Q∗l|Xl|.
Then, since Ql∼ κ00Q∗l with a suitable κ006= 0,
n−1
X
l=k+1
QlXl= O
n−1
X
l=k+1
Q∗l|Xl|
!
= O
|Xn−1|
(n − 1)2µ −|Xk| k2µ
. Finally, by (A.16), (A.17), (A.18) and the last above relations, we obtain for x 6= 0
Gk+1,n−1(x, q) = C(x, q) Fk+1,n−1ν (q)
(n − 1)µ −Fk+1,n−1γ (x) kµ
!
+ O |Fk+1,n−1ν (q)|
n2µ +|Fk+1,n−1γ (x)|
k2µ
! .
A.4. Asymptotic behavior of C
mJ (I)0,n−1P
j∈J
Y
j,m0Recalling (Aletti, Crimaldi and Ghiglietti, 2019, Equation (5.37)) and taking into account the fact that in all the considered cases with 1 ∈ J , i.e. (ii), (iii) and (v), we have 1 /∈ I1, we get
CmJ (I)0,n
= O(Cn11) + O(Cn21) + O(Cn22), where
Cn11 := X
j∈J, j6=1
1{1∈Ij}|Fmγ
0,n−1(cαj)| , Cn21 := X
j∈J, j6=1
1{2∈Ij}|Gm0,n−1(cαj, q)| ,
Cn22 := X
j∈J
1{2∈Ij}|Fmν0,n−1(q)| .
Using (A.5) and denoting by a∗ the real part of α∗:= 1 − λ∗, it is immediate to see that
Cn11= X
j∈J, j6=1
1{1∈Ij}
O
exp
−ca∗n1−γ 1 − γ
if 1/2 < γ < 1
O(n−ca∗) if γ = 1
and
Cn22=X
j∈J
1{2∈Ij}
O
exp
−qn1−ν 1 − ν
if 1/2 < ν < 1
O(n−q) if ν = 1.
For the term Cn21, we apply LemmaA.4so that we get:
Case ν < γ We have Gm0,n−1(cαj, q) = O n−(γ−ν)|Fmν0,n−1(q)| + |Fmγ0,n−1(cαj)| by means of LemmaA.4 and so
Cn21= X
j∈J, j6=1
1{2∈Ij}O
n−(γ−ν)|Fmν0,n−1(q)| + |Fmγ0,n−1(cαj)| ,
where, as above, by (A.5), we have |Fmν0,n−1(q)| = O exp
−qn1−ν1−ν
and
|Fmγ
0,n−1(cαj)| =
O
exp
−ca∗n1−γ 1 − γ
if 1/2 < γ < 1
O(n−ca∗) if γ = 1.
Case ν > γ We have Gm0,n−1(cαj, q) = O n−(ν−γ)|Fmν0,n−1(q)| + |Fmγ
0,n−1(cαj)| by means of LemmaA.4 and so
Cn21= X
j∈J, j6=1
1{2∈Ij}O
n−(ν−γ)|Fmν0,n−1(q)| + |Fmγ
0,n−1(cαj)| ,
where, as above, by (A.5), we have |Fmγ
0,n−1(cαj)| = O exp
−ca∗ n1−γ1−γ
and
|Fmν0,n−1(q)| =
O
exp
−qn1−ν 1 − ν
if 1/2 < ν < 1
O(n−q) if ν = 1.
Case ν = γ Assuming q 6= cαj for all j ≥ 2, by LemmaA.4, we have1 Gm0,n−1(cαj, q) = O |Fmγ
0,n−1(q)| +
|Fmγ0,n−1(cαj)| and so
Cn21= X
j∈J, j6=1
1{2∈Ij}O |Fmγ0,n−1(q)| + |Fmγ0,n−1(cαj)| ,
where, as above, by (A.5), we have for x = q or x ∈ {cαj : j ∈ J, j 6= 1}
|Fmγ0,n−1(x)| =
O
exp
−ax
n1−γ 1 − γ
if 1/2 < ν = γ < 1
O(n−ax) if ν = γ = 1
and so, setting x∗:= min{q, ca∗}, we can write
Cn21= X
j∈J, j6=1
1{2∈Ij}
O
exp
−x∗n1−γ 1 − γ
if 1/2 < ν = γ < 1
O(n−x∗) if ν = γ = 1.
Summing up, taking into account the conditions ca∗ > 1/2 when γ = 1 and q > 1/2 when ν = 1, we can conclude that in all the six cases (i)-(vi) we have tn(J (I))
CmJ (I)0,n−1
→ 0 and so tn(J (I))CmJ (I)0,n−1X
j∈J
Yj,m0
−→ 0.a.s.
A.5. Asymptotic behavior of P
n−1k=m0
ρ
J (I)k+1,n−1Recall (Aletti, Crimaldi and Ghiglietti,2019, Equation (5.39)) and the fact that ∆RY,k+1= (∆RZ,k+1, ∆RN,k+1)>, where, by (Aletti, Crimaldi and Ghiglietti, 2019, Assumption 2.2), we have |∆RZ,k+1| = O(k−2γ) and
|∆RN,k+1| = O(k−2ν). Then, taking into account the fact that in all the considered cases with 1 ∈ J , i.e. (ii), (iii) and (v), we have 1 /∈ I1, we get
n−1
X
k=m0
ρJ (I)k+1,n−1
= O(ρ11n ) + O(ρ21n ) + O(ρ22n ),
1If there exists j ≥ 2 such that q = cαj, we have to consider the other asymptotic expression given in LemmaA.4.
where
ρ11n := X
j∈J, j6=1
1{1∈Ij} n−1
X
k=m0
k−2γ|Fk+1,n−1γ (cαj)| ,
ρ21n := X
j∈J, j6=1
1{2∈Ij} n−1
X
k=m0
k−2γ|Gk+1,n−1(cαj, q)| ,
ρ22n :=X
j∈J
1{2∈Ij} n−1
X
k=m0
(k−2γ+ k−2ν)|Fk+1,n−1ν (q)|.
Using Lemma A.2(with β = 2γ > 1, e = 1 and δ = γ), we get
ρ11n = X
j∈J, j6=1
1{1∈Ij}
O (n−γ) if 1/2 < γ < 1, O n−ca∗
if γ = 1 and 1/2 < ca∗< 1, O n−1ln(n)
if γ = 1 and ca∗= 1, O n−1
if γ = 1 and ca∗> 1.
For ρ22n , we observe that we have k−2γ = O(k−2ν) when ν ≤ γ and k−2ν = O(k−2γ) when ν > γ. Therefore, using LemmaA.2 (with e = 1 and δ = ν and β = 2ν > 1 if ν ≤ γ and β = 2γ > 1 if ν > γ), we obtain for the case ν ≤ γ
ρ22n =X
j∈J
1{2∈Ij}O
n−1
X
k=m0
k−2ν|Fk+1,n−1ν (q)|
!
=X
j∈J
1{2∈Ij}
O (n−ν) if 1/2 < ν < 1,
O (n−q) if ν = 1 and 1/2 < q < 1, O n−1ln(n)
if ν = 1 and q = 1, O n−1
if ν = 1 and q > 1 , and for the case ν > γ
ρ22n =X
j∈J
1{2∈Ij}O
n−1
X
k=m0
k−2γ|Fk+1,n−1ν (q)|
!
=X
j∈J
1{2∈Ij}
O n−2γ+ν
if 1/2 < ν < 1,
O (n−q) if ν = 1 and 1/2 < q < 2γ − 1, O (n−qln(n)) if ν = 1 and q = 2γ − 1 > 1/2, O n−2γ+1
if ν = 1 and q > max{1/2, 2γ − 1} .
(A.19)
For the term ρ21n , we apply LemmaA.2and Lemma A.4so that we get:
Case ν < γ We have Gk+1,n−1(cαj, q) = O n−(γ−ν)|Fk+1,n−1ν (q)| + k−(γ−ν)|Fk+1,n−1γ (cαj)| by means of LemmaA.4, and so we get
ρ21n = X
j∈J j6=1
1{2∈Ij}O n−(γ−ν)
n−1
X
k=m0
1
k2γ|Fk+1,n−1ν (q)| +
n−1
X
k=m0
1
k3γ−ν|Fk+1,n−1γ (cαj)|
! ,
where, by LemmaA.2, the first term is O(n−3γ+2ν), while for the second term we have
n−1
X
k=m0
1
k3γ−ν|Fk+1,n−1γ (cαj)| =
O n−2γ+ν
if 1/2 < γ < 1, O n−ca∗
if γ = 1 and 1/2 < ca∗ < 2 − ν, O n−2+νln(n)
if γ = 1 and ca∗= 2 − ν, O n−2+ν
if γ = 1 and ca∗> 2 − ν .
Case ν > γ We have Gk+1,n−1(cαj, q) = O n−(ν−γ)|Fk+1,n−1ν (q)| + k−(ν−γ)|Fk+1,n−1γ (cαj)| by means of LemmaA.4, and so we get
ρ21n = X
j∈J, j6=1
1{2∈Ij}O n−(ν−γ)
n−1
X
k=m0
1
k2γ|Fk+1,n−1ν (q)| +
n−1
X
k=m0
1
kγ+ν|Fk+1,n−1γ (cαj)|
! ,
where, by LemmaA.2, the second term is O(n−ν), while the sum in the first term has the asymptotic behavior given in (A.19).
Case ν = γ Assuming q 6= cαjfor all j ≥ 2, by LemmaA.4, we have2Gk+1,n−1(cαj, q) = O |Fk+1,n−1γ (q)| +
|Fk+1,n−1γ (cαj)|, and so we get
ρ21n = X
j∈J, j6=1
1{2∈Ij}O
n−1
X
k=m0
1
k2γ|Fk+1,n−1γ (q)| +
n−1
X
k=m0
1
k2γ|Fk+1,n−1γ (cαj)|
! ,
where, by LemmaA.2, we have for x = q or x ∈ {cαj: j ∈ J, j 6= 1}
n−1
X
k=m0
1
k2γ|Fk+1,n−1γ (x)| =
O (n−γ) if 1/2 < ν = γ < 1,
O (n−ax) if ν = γ = 1 and 1/2 < ax< 1, O n−1ln(n)
if ν = γ = 1 and ax= 1, O n−1
if ν = γ = 1 and ax> 1 and so, setting x∗:= min{q, ca∗}, we can write
ρ21n = X
j∈J, j6=1
1{2∈Ij}
O (n−γ) if 1/2 < ν = γ < 1, O n−x∗
if ν = γ = 1 and 1/2 < x∗ < 1, O n−1ln(n)
if ν = γ = 1 and x∗= 1, O n−1
if ν = γ = 1 and x∗> 1 .
Summing up, taking into account the conditions ca∗> 1/2 when γ = 1 and q > 1/2 when ν = 1, from the asymptotic behavior given above we easily obtain that in all the cases (i)-(v) we have tn(J (I))
Pn−1
k=m0ρJ (I)n,k → 0 a.s. and so
tn(J (I))
n−1
X
k=m0
ρJ (I)n,k −→ 0 .a.s. (A.20)
In the case (vi), the evaluation of the asymptotic behavior given in (A.19) for the term ρ22n is not enough in order to conclude that tn(J (I))ρ22n → 0. Therefore, we need a better evaluation, that we can get applying Lemma A.2in a different way. Indeed, in the case (vi), taking u > 1 and applying Lemma A.2with e = u,
2If there exists j ≥ 2 such that q = cαj, we have to consider the other asymptotic expression given in LemmaA.4.
δ = ν and β = 2γu > 1, we find
tn(J (I))ρ22n u
= nuν/2O
n−1
X
k=m0
k−2γu|Fk+1,n−1ν (q)|u
!
= nuν/2
O n−2γu+ν
if 1/2 < ν < 1,
O (n−qu) if ν = 1 and 1/2 < q < 2γ − u−1, O (n−quln(n)) if ν = 1 and q = 2γ − u−1 > 1/2, O n−2γu+1
if ν = 1 and q > max{1/2, 2γ − u−1} .
Hence, from the above relations we get that it is possible to find u > 1 large enough such that (tn(J (I))ρ22n)u→ 0, that trivially implies tn(J (I))ρ22n → 0. Therefore also in the case (vi), we can conclude that (A.20) holds true.
A.6. Computation of the limit d
j1(i1),j2(i2)Recall that we have dj(1)k,n =
(
rbk−1 for j = 1
rbk−1Fk+1,n−1γ (cαj) for j ≥ 2 and
dj(2)k,n =
( (qbk,k−brk−1) Fk+1,n−1ν (q) for j = 1
λjrbk−1Gk+1,n−1(cαj, q) + (qbk,k−brk−1g(λj)) Fk+1,n−1ν (q) for j ≥ 2 , where, for each j ≥ 2, we have g(λj) = λj when ν < γ, while g(λj) = 0 when ν ≥ γ.
Here, for each of the six cases (i) − (vi) listed in the statement of (Aletti, Crimaldi and Ghiglietti,2019, Lemma 5.1), we compute the limit
dj1(i1),j2(i2) = lim
n tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2).
For all the computations, we make the assumptions stated in (Aletti, Crimaldi and Ghiglietti,2019, Section 2) and we use LemmaA.3and LemmaA.4.
Case (i) Take ν < γ, j1, j2∈ {2, . . . , N } and i1= i2= 1. We have
limn tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2)= lim
n nγ
n−1
X
k=m0
br2k−1Fk+1,n−1γ (cαj1)Fk+1,n−1γ (cαj2)
= c2lim
n nγ
n−1
X
k=m0
k−2γFk+1,n−1γ (cαj1)Fk+1,n−1γ (cαj2)
= c2
c(αj1+ αj2) −1{γ=1}
.
Case (ii) Take ν < γ, j1, j2∈ {1, . . . , N } and i1= i2= 2. For j1= j2= 1, we have
limn tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2)= lim
n nν
n−1
X
k=m0
(qbk,k−brk−1)2Fk+1,n−1ν (q)2
= lim
n nν
n−1
X
k=m0
bqk,k2 Fk+1,n−1ν (q)2
= q2lim
n nν
n−1
X
k=m0
k−2νFk+1,n−1ν (q)2= q 2.
(Note that the above second equality is due to the fact that some terms are o(n−ν) and so we can cancel them.) Similarly, for the cases j1≥ 2, j2≥ 2 and j1= 1, j2≥ 2 and j1≥ 2, j2= 1, using LemmaA.4, which allows us to replace in the computation of the desired limit the quantity Gk+1,n−1(cαj, q) by
−cαj q
Fk+1,n−1ν (q)
(n − 1)γ−ν −Fk+1,n−1γ (cαj) kγ−ν
! , and removing the terms which are o(n−ν), we obtain
lim
n tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2)= lim
n nν
n−1
X
k=m0
bqk,k2 Fk+1,n−1ν (q)2
= q2lim
n nν
n−1
X
k=m0
k−2νFk+1,n−1ν (q)2= q 2.
Case (iii) Take ν = γ, j1, j2 ∈ {1, . . . , N } and i1, i2 ∈ {1, 2} with ih 6= 1 if jh = 1. Recall that we are assuming q 6= cαj for all j ≥ 23. Therefore, for j1= j2= 1 and i1= i2= 2, we have
limn tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2)= lim
n nγ
n−1
X
k=m0
(qbk,k−rbk−1)2Fk+1,n−1γ (q)2
= lim
n (q − c)2nγ
n−1
X
k=m0
1
k2γFk+1,n−1γ (q)2= (q − c)2 2q −1{γ=1}
.
For j1= 1, j2≥ 2, i1= 2 and i2= 1, we have limn tn(J (I))2
n−1
X
k=m0
djk,n1(i1)djk,n2(i2)=
limn nγ
n−1
X
k=m0
(qbk,k−brk−1) Fk+1,n−1γ (q)rbk−1Fk+1,n−1γ (cαj2) =
(q − c)c lim
n nγ
n−1
X
k=m0
1
k2γFk+1,n−1γ (q)Fk+1,n−1γ (cαj2) = c(q − c) cαj2+ q −1{γ=1}
.
By symmetry, for j1≥ 2, j2= 1, i1= 1 and i2= 2, we have dj1(i1),j2(i2)= c(q − c)
cαj1+ q −1{γ=1}
.
3In the case q = cαjfor some j ≥ 2 the computations are similar, but we have to consider the other asymptotic expression given in LemmaA.4.